I'm reading through the following paper: Monte Carlo simulation of non relativistic electron scattering by W. Williamson and G. C. Duncan.
In the following paragraph, I want to know how to arrive at $(2)$ from $(1)$. (It's basically a change of units) :
Bethe has derived an expression which gives the kinetic energy lost by a non-relativistic electron as it traverses a path of length $ds$ in matter. We assume that the energy lost per unit path length by the electron is given by the Bethe formula: $$\frac{\mathrm{d}T}{\mathrm{d}s} = -\frac{2\pi e^4}{T} NZ \ln\left(\frac{2T(\mathrm{eV})}{11.5Z}\right)\tag{1}$$
In Eq. (I), $T$ is the kinetic energy of the electron, $e$ is the electron charge, $N$ is the number of target atoms per $cm^3$, and $Z$ is the atomic number. For calculations it is convenient to express the energy loss in the units of $keV \over\mu m$ and in terms of the atomic weight and density of the target material. In these units, Eq. $(1)$ becomes: $$ \frac{\mathrm{d}T}{\mathrm{d}s} = -7.83 \left(\frac{\rho Z}{AT}\right) \ln\left(\frac{174T}{Z}\right) \left(\frac{\mathrm{keV}}{\mu\mathrm{m}}\right) \tag{2} $$ where in equation $(2)$ $\rho$ ($\mathrm{g\over cm^3}$) is the density of the target, $A$ ($\mathrm{g}$) is the atomic weight of the target, and $T$ ($\mathrm{keV}$) is the electronic kinetic energy.
The problems is with the numerical coefficient $7.83$, where the numerical values of $e$ and $\pi$ and $N=\frac{\rho N_A}{A}$ are substituted in equation $(1)$. In CGS units, $e=4.80320427 \times 10^{−10} \,\,\,(Fr)$, which gives a wrong value when substituted in $(1)$. (I found out that the correct result can be obtained, if you divide it by $1.602\times 10^{-19}$, which is electronic charge in SI).